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I have recently started learning Algebraic Topology, and more specifically, homotopy theory. That is where I encountered this question:

Let $R$ be the set $\mathbb{R}$ of real numbers, with the co-finite topology. Show that $R$ is contractible.

I already have some knowledge of point-set topology, the definition of homotopy, homotopy equivalence and the meaning of a contractible space.

The idea I used to approach the problem was to try and show that the identity map on $R$ was null homotopic. I considered the identity map $id:R\to R$ and $c:R\to R$, defined as $c(x)=x_0$ for all $x\in R$, where $x_0\in R$. I tried to check if the function $H:R\times\mathbb{I}\to R$ (where $\mathbb{I}=[0,1]$), given by $H(x,t)=(1-t)x+tx_0$, was continuous in the product topology by checking if the pre-images of closed sets are closed, but that is where I am stuck. I am finding it hard to do that.

Besides that, I found a result on the following webpage that might be helpful, albeit requiring a different approach. If it is possible to show that $R$ is path-connected, then that result suggests that it would be contractible. I am yet to perfectly understand the proof to it, but it could be applied too. Here's the link to it: A strange contractible space.

Any help would be much appreciated.

Vishal Agarwal
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  • See https://math.stackexchange.com/questions/514578/what-is-the-homotopy-type-of-the-affine-space-in-the-zariski-topology/684947#684947 – Moishe Kohan Jun 08 '21 at 05:01
  • @MoisheKohan I am actually not aware of the Zariski topology. Is it exactly the same as the co-finite topology? Also, how, in that answer, the OP said that there would exist a bijection $\phi: \mathbb{C}^{1} \times (0, 1) \simeq \mathbb{C}^{1}$ and thus the pre-images of points would be closed, isn't really clear to me. – Vishal Agarwal Jun 08 '21 at 05:18
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    Yes, in your case Zariski topology is the same as the cofinite topology. If you have a bijection, preimages of points are points, so they are closed. – Moishe Kohan Jun 08 '21 at 05:28
  • @MoisheKohan Okay, I see what you are getting at. I now need to convince myself that there does, in fact, exist such a bijection between that product space and the $\mathbb{C}^1$ space. – Vishal Agarwal Jun 08 '21 at 06:43
  • $X$ and $X^2$ have the same cardinality if $X$ is infinite. From this, you see that ${\mathbb R}$ and ${\mathbb R}\times (0,1)$ have the same cardinality. – Moishe Kohan Jun 08 '21 at 12:28
  • @MoisheKohan Right, I do remember having read a way to find a bijection between $\mathbb{R}$ and $\mathbb{R}^2$. Moreover, we can have a bijection between (0,1) and $\mathbb{R}$. Thank you. – Vishal Agarwal Jun 08 '21 at 12:37

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In your linked blogpost there is the lemma

Let $X$ have the cofinite topology; then if $X$ is path-connected, then $X$ is contractible.

And $\Bbb R$ in the cofinite topology is path-connected: if $x \neq y$ in $\Bbb R$ then $p: [0,1] \to \Bbb R, p(t)=tx+(1-t)y$ is injective and thereby automatically continuous when the codomain has the cofinite topology (inverse images of closed sets are closed!), as $[0,1]$ is metric hence $T_1$.

So the proof in the blog applies and $\Bbb R$ in the cofinite topology is contractible.

Henno Brandsma
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